sinx cos^3x cosx sin^3x
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(1)化简可得f(x)=4sinx(cosxcosπ3-sinxsinπ3)+3=2sinxcosx-23sin2x+3=sin2x+3cos2x…(2分)=2sin(2x+π3)…(4分)所以T=2
fx=-√3cos2x-sin2x=-2sin(2x+π/3)所以最小正周期为πf'x=-4cos(2x+π/3),f'x>0时递增x在(π/12,π/3)上递增f'x=0,x=π/12.极小值f(π
请问,会不会多了一个cosx?
令sinxdx=-d(cosx)t^3/(1+t^2)dt=[(t^3+t)-t]/(1+t^2)*dt=t-t/(1+t^2)t^2/2-1/2*ln(1+t^2)+Ccosx^2/2-ln(1+c
原式=2sinxcos(x+π/3)+√3cos²x+sinxcosx=2sinxcos(x+π/3)+cosx(√3cosx+sinx)=2sinxcos(x+π/3)+2cosx·sin
f(x)=2(cosx)^2-2√3sinxcosx-1=(cos2x+1)-√3sin2x-1=cos2x-√3sin2x=2cos(2x+π/6)周期T=2π/│ω│=2π/2=π因为y=cosx
F(X)=5√3cos²x+√3sin²x-4sinxcosx=√3cos²x+√3sin²x+4√3cos²x-4sinxcosx=√3+2√3(c
才5分==再问:提高了再答:1、π;5/2,1/22、-π/12再问:第二个问能详解一下么?谢再答:奇函数的话就意味着有一个对称中心(0,0),这时是最小的
f(x)=(1+cos2x)/2+(√3/2)sin2x+3/2=(√3/2)sin2x+(1/2)cos2x+2=sin2xcos(π/6)+cos2xsin(π/6)+2=sin(2x+π/6)+
(1)最简单的方法是用“积化和差”公式2sinαcosβ=sin(α+β)+sin(α-β)原式=2×2sinxcos(x+π/3)=2[sin(x+x+π/3)+sin(x-x-π/3)]=2[si
y'=cosx-3sin²xcosx
=-1/2cos2x+根号3)/2sin2x+3/2=sin(2x-£/6)+3/2
y=2sin²x-√3sinxcosx+cos²x=sin²x-√3sinxcosx+(sin²x+cos²x)=(1-cos2x)/2-√3/2*s
f(x)=5√3(cosx)^2+√3(sinx)^2-4sinxcosx=5√3*(1+cos2x)/2+√3*(1-cos2x)/2-2sin2x=2√3cos2x-2sin2x+3√3=4(√3
y=2sinxcos^2x/(1+sinx)=2sinx﹙1-sin²x﹚/(1+sinx)=2sinx﹙1-sinx﹚=-2﹙sinx-½﹚²+½y=sin^
y=2√3*sinxcosx+2cos^2x=√3sin2x+cos2x+1=sin(2x+π/6)+1∴最小正周期:t=2π/2=π
正在解答再问:答案呢我问你再答:放心,包正确再答:正在解答啊再问:答案给我就给好评再问:要全面再答:再答:再做第二问再答:合作愉快再答:把横坐标变为原来的二分之一再问:亮一点再答:图像向左平移六分之派
(1)∵f(x)=2sinxcos(π2-x)-3sin(π+x)cosx+sin(π2+x)cosx=2sin2x+3sinxcosx+cos2x=32sin2x-12cos2x+32=sin(2x
解原式=2sinxcos(x+π/3)+根号3cos的平方x+1/2sin2x=2sinxcos(x+π/3)+根号3cos的平方x+sinxcosx=2sinxcos(x+π/3)+cosx(根号3
9.已知函数f(x)=sinx*2+2√3sinxcos+3cos*2x(2)已知f(a)=3,且a∈(0,π),求a的值f(x)=sin^x+2√3sinxcosx+3cos^2x=(sin^2x+